Physics · Gravitation and the universe · Grade 10

Kepler's laws

What path does a planet follow around the Sun, and at what speed?

The situation

In the 15th century, at the observatory in Samarkand, Mirzo Ulugh Beg and his pupils observed the Sun, the Moon and the stars with a huge meridian arc of radius about 40 m. Every day at noon, sunlight passed through a slit at the centre of the arc and fell on graduated marble rails — this spot moves up and down through the year. Such precise measurements later became the foundation of celestial mechanics: at the beginning of the 17th century Johannes Kepler discovered the three laws of planetary motion. Along what path and at what speed does a planet move around the Sun?

Topic

Kepler's laws. Mirzo Ulugh Beg's observatory and the motion of the planets

Key ideas

Ellipse and focus
Law of equal areas
T² = a³

Formulas

r = \frac{a (1 - e^{2})}{1 + e \cos θ}, r_{m i n} = a (1 - e), r_{m a x} = a (1 + e)

Kepler's first law: an orbit is an ellipse, with the Sun at one of its foci.

\frac{Δ S}{Δ t} = \frac{L}{2 m} = const

Second law: the radius vector sweeps out equal areas in equal times (angular momentum $L$ is conserved).

T^{2} = \frac{4 π^{2}}{G (M_{⊙} + m)} a^{3} \Rightarrow \frac{T^{2}}{a^{3}} \approx 1 \frac{{yr}^{2}}{{AU}^{3}} (m ≪ M_{⊙})

Third law: the period depends only on the semi-major axis, not on the eccentricity. In its exact form the sum of the masses of the two bodies appears.

v^{2} = G (M_{⊙} + m) (\frac{2}{r} - \frac{1}{a}), μ = \frac{M_{⊙} m}{M_{⊙} + m}

The vis-viva equation (for relative speed) and the reduced mass $μ$ : the two-body problem is reduced to a one-body problem.

h_{n o o n} = 90^{\circ} - φ + δ

The observatory: the altitude of the Sun at noon (for Samarkand $φ \approx {39.7}^{\circ}$ , $δ$ is the Sun's declination, $\pm {23.4}^{\circ}$ ).

Theory

First law: the ellipse

Under the gravitational force $F = G M m / r^{2}$ a closed orbit is an ellipse, and the Sun stands not at its centre but at a focus. The eccentricity $e$ expresses how elongated it is: $e = 0$ is a circle, $e \to 1$ a very elongated ellipse (for Halley's Comet $e \approx 0.967$ ). The nearest point is the perihelion $q = a (1 - e)$ , the farthest the aphelion $Q = a (1 + e)$ . The second focus $F_{2}$ is empty.

Second law: equal areas

The gravitational force is always directed towards the Sun (a central force), so it does not “twist” the planet — the angular momentum $L = m r v_{⟂}$ is conserved. As a result the radius vector sweeps out equal areas in equal times: at perihelion the sector is short and wide, at aphelion long and thin. That is why a planet speeds up as it approaches the Sun. Switch on “Sectors” — the areas are compared in a bar chart.

Third law and the logarithmic graph

From Newton's law of gravitation, $T^{2} = 4 π^{2} a^{3} / G M_{⊙}$ . In years and AU this is simply $T^{2} = a^{3}$ . The values of $a$ for the planets range from 0.39 to 30 AU — an 80-fold difference, so the graph is drawn on a logarithmic scale: all the planets and the orbits you measure lie on one straight line. Notice: if you change $e$ , the period does not change. This is a consequence of mechanical similarity: if the potential is $U \propto r^{k}$ , then when all lengths are scaled by $λ$ the equation of motion is unchanged if time is stretched by a factor of $λ^{1 - k / 2}$ . For gravity $k = - 1$ , so $T \propto a^{3 / 2}$ ; for a spring ( $k = 2$ ) the period does not depend on the amplitude.

The two-body problem and reduced mass

In fact the Sun is not stationary either: both bodies orbit their common centre of mass. For the relative vector $\vec{r} = {\vec{r}}_{2} - {\vec{r}}_{1}$ the equation becomes a one-body problem: in place of the mass stands the reduced mass $μ = M m / (M + m)$ , and in place of the mass of the attracting centre stands $M + m$ ; hence $G (M + m)$ in the third law. The Sun traces a small ellipse with the same period as the planet, always on the opposite side from it, with semi-major axis $a m / (M + m)$ . For Jupiter this is ≈ 0.005 AU — slightly more than the radius of the Sun: the centre of mass lies just outside the Sun's surface. This “wobble” of the Sun (≈ 12.5 m/s) is one of the methods of finding planets around distant stars. In the “Two bodies” section increase $m$ : at $m = M$ two stars trace identical ellipses.

Effective potential, the virial theorem, Hamilton–Jacobi

In a central field the angular momentum $L$ is conserved and the radial motion becomes one-dimensional: $E = \frac{1}{2} μ {\dot{r}}^{2} + U_{e f f} (r)$ , $U_{e f f} = \frac{L^{2}}{2 μ r^{2}} - \frac{G M m}{r}$ . The first term is the centrifugal barrier. The body can move only in the region where $E \geq U_{e f f}$ : at $E < 0$ it oscillates in a “well” between perihelion and aphelion (an ellipse), and at $E \geq 0$ it goes off to infinity. See the well and the line $E$ on the graph. Virial theorem: for $U \propto 1 / r$ the averages over a period satisfy $2 ⟨ K ⟩ = - ⟨ U ⟩$ , i.e. $⟨ K ⟩ = - E$ — this is checked in the table of results after each revolution. In the Hamilton–Jacobi method the action function separates in the variables: $S = - E t + L φ + \int \sqrt{2 μ (E - U_{e f f} (r))} d r$ — under the root is the same effective potential; from it follows the orbit equation $r (φ)$ (the general method is in the “Hamiltonian mechanics” experiment).

Model and observatory

The simulation integrates Newton's laws for the relative vector directly (the velocity Verlet method, with the step made smaller as the planet approaches the Sun); at $m = 0$ the Sun is stationary, and the influence of the other planets is not taken into account. The Cavendish experiment that measured the gravitational constant $G$ is in the “Law of universal gravitation” experiment. Kepler's laws are not built in here — they come out of the calculation by themselves. The observatory model is a simplified reconstruction; the height of the spot on the meridian arc is calculated from the Sun's annual motion using the formula $h = 90^{\circ} - φ + δ$ . Time is sped up: at 1× 1 s = 0.1 year.

Did you know?

Ulugh Beg's “Zij-i Sultani” tables give the positions of 1018 stars. He measured the length of the sidereal year as 365 days 6 hours 10 min 8 s — only about a minute more than the modern value.
Kepler published his first two laws in 1609 in “New Astronomy”, and the third in 1619 in “Harmony of the World”. He used the positions of Mars measured by Tycho Brahe.

Inside the experiment

The experiment starts from a real-life situation: you predict the result, change the controls, run the experiment and draw a conclusion. Inside are missions, theory, a quiz and a guided tour in which the camera moves and explains the key points.

More experiments in this section

Scales of the universe — Grade 9
Newton's cannon: an artificial satellite — Grade 10
Special relativity — Grade 10
Law of universal gravitation — Grade 10
Relativistic dynamics — Grade 10
Physics of stars — Year 1
The expanding universe — Year 1
Black holes — Year 4